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a) ( x + 2 )( x2 - 2x + 4 ) - ( 18 + x3 )
= x3 + 8 - 18 - x3 = -10
b) ( 2x - y )( 4x2 + 2xy + y2 ) - ( 2x + y )( 4x2 - 2xy + y2 )
= 8x3 - y3 - ( 8x3 + y3 )
= 8x3 - y3 - 8x3 - y3 = -2y3
c) ( x - 3 )( x + 3 ) - ( x + 5 )( x - 1 )
= x2 - 9 - ( x2 + 4x - 5 )
= x2 - 9 - x2 - 4x + 5 = -4x - 4
d) ( 3x - 2 )2 + ( x + 1 )2 + 2( 3x - 2 )( x + 1 )
= ( 3x - 2 + x + 1 )2
= ( 4x - 1 )2
a) (x+3)(x^2-3x+9)-(54+x^3)
= x^3- 3x^2+9x+3x^2-9x+27-54-x63
= -27
b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)
= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]
= [(2x)3^3+ y^3] – [(2x)^3 – y^3]
= (2x)^3 + y^3 – (2x)^3 + y^3
= 2y^3
a)(x+3)(X^2-3x+9)-(54+x^3)
= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)
= 27- 54
= -27
b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)
= \((2x)^3\) + \(y^3\) - [\((2x)^3\) - \(y^3\) ]
= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)
= \(2y^3\)
b) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)+\left(2x+y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2+4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(8x^2+2y^2\right)\)
\(=\left(2x+y\right)\left(4x+y\right).2xy\)
\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)
\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)
\(=-5x-27\)
\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y+z-x-y\right)^2\)
\(=z^2\)
a)
=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\)
=-5x-19
b)
=\(8x^3+y^3-8x^3+y^3\)
=\(2y^3\)
c)
=(x+y+z-x-y)\(^2\) +x+y
=\(z^2+x+y\)
hc tốt
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
\(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=x^2-6x+9-x^2-4x-4\)
\(=-10x+5\)
\(\left(4x^2-2xy+y^2\right)\left(2x-y\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right)\left(4x^2-2xy+y^2-4x^2-2xy-y^2\right)\)
\(=\left(2x-y\right)\cdot\left(-4xy\right)\)
a,\(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=x^2-6x+9-x^2-4x-4\)
\(=-10x+5\)
b, \(\left(4x^2-2xy+y^2\right).\left(2x-y\right)-\left(2x-y\right).\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right).\left(4x^2-2xy+y^2-4x^2-2xy-y^2\right)\)
\(=\left(2x-y\right).\left(-4xy\right)\)
1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)
b, (a+b) ^3 - ( a - b)^3- 2 b^3
= ( a +b -a +b) [ ( a+ b)^2 + (a+b)(a-b) + (a-b)^2] - 2b^3
= 2b( a^2 + 2ab+ b^2 + a^2 - b^2 + a^2 - 2ab+ b^2 ) - 2b^3
= 2b ( 3 a^2 + b^2) - 2b^3
= 2b ( 3a^2 + b^2 - b^2)
= 2b.3a^2
=6a^2b
a,=(x+2).(x^2-2x+2^2)-18-x^3
=x^3 + 2^3 - 18 -x^3=(x^3-x^3)+(8-18) = -10
b, =(2x-y).((2x)2+2xy +y2) - (2x +y).((2x)^2-2xy +y^2)
=(2x)3-y3- (2x)3-y3= -2 .y3
2 ý này áp dụng HĐT : x3+y3=(x+y).(x2-xy+y2)
x3-y3=(x-y).(x2+xy+y2)